Limit and integral

As the limits range from 0 to 1.So $x$ is a number between 0 and 1.

$\displaystyle \lim_{n\to \infty} x^{n} \rightarrow 0$

So the integral reduces to $\displaystyle \int^{1}_{0} |-x|.dx$

$\frac{x^2}{2}$ limits from 0 to 1

$=0.5$

U can also try this

See, the integration is inside the interval $\left ( 0,1 \right )$ . But in this interval we have

$0 < x < 1 \;\; \Rightarrow \;\; 0 < x^n < x$

And, therefore, as $|x| > 0$ for $x$ in our interval,

$x^n - |x| < 0 \;\; \Rightarrow \;\; |x^n - |x|| = x - x^n$

Therefore the integral becomes

$\int_{0}^{1} |x^n - |x|| \; \mathrm{d}x = \int_{0}^{1} \left (x - x^n \right ) \; \mathrm{d}x = \frac{1}{2} - \frac{1}{n+1}$

And

$\lim_{n\rightarrow \infty} \left (\frac{1}{2} - \frac{1}{n+1} \right ) = 0.5$